Download HW_2_AMS 570 Q.1. Suppose that and are random variables with

HW_2_AMS 570
Q.1. Suppose that 𝒀𝟏 and 𝒀𝟐 are random variables with joint pdf
𝟖𝒚 𝒚 , 𝟎 < 𝒚𝟏 < 𝒚𝟐 < 𝟏
𝒇𝒀𝟏 ,𝒀𝟐 (𝒚𝟏 , 𝒚𝟐 ) = { 𝟏 𝟐
𝟎,
𝐨𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞.
Find the pdf of 𝑼𝟏 = 𝒀𝟏 /𝒀𝟐 .
We are given that the joint pdf of 𝑋 and 𝑌is𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 8𝑦1 𝑦2 , 0 < 𝑦1 < 𝑦2 < 1.
Now we need to introduce a second random variable 𝑈2 which is a function of 𝑋 and𝑌.
We wish to do this in such a way that the resulting bivariate transformation is one-to-one
and our actual task of finding the pdf of 𝑈1 is as easy as possible. Our choice of 𝑈2 is of
course, not unique. Let us define𝑈2 = 𝑌2 . Then the transformation is, (using𝑢1 , 𝑢2 , 𝑦1 , 𝑦2 ,
since we are really dealing with the range spaces here).
𝑦1 = 𝑢1 ∗ 𝑢2
𝑦2 = 𝑢2
From it, we find the Jacobian,
𝑢
J=| 2
0
𝑢1
| = 𝑢2
1
To determineℬ, the range space of 𝑈1 and𝑈2 , we note that
0 < 𝑦1 < 𝑦2 < 1 ⇒ 0 < 𝑢1 ∗ 𝑢2 < 𝑢2 < 1, equivalently,
𝑢1 > 0
𝑢1 < 1
𝑢2 > 0
𝑢2 < 1
So ℬ is as indicated in the diagram below.
1
𝑓𝑈1,𝑈2 (𝑢1 , 𝑢2 ) = 8 ∗ (𝑢1 ∗ 𝑢2 ) ∗ 𝑢2 ∗ 𝑢2 = 8𝑢1 𝑢23 ,
Thus, the marginal pdf of 𝑈1 is obtained by integrating 𝑓𝑈1,𝑈2 (𝑢1 , 𝑢2 ) with respect to𝑢2 ,
giving
1
𝑓𝑈1 (𝑢1 ) = ∫0 8𝑢1 𝑢23 𝑑𝑢2 = 2 ∗ 𝑢1 , 0 < 𝑢1 < 1 .
𝟏 𝟐 𝒙
2.3 Suppose X has the geometric pmf 𝒇𝑿 (𝒙) = 𝟑 (𝟑) , 𝒙 = 𝟎, 𝟏, 𝟐, … Determine the
𝑿
probability distribution of 𝒀 = 𝑿+𝟏. Note that here both X and Y are discrete
random variables. To specific the probability distribution of Y, specify its pmf.
𝑦
𝑋
𝑦
1
𝑃(𝑌 = 𝑦) = 𝑃 (𝑋+1 = 𝑦) = 𝑃 (𝑋 = 1−𝑦) = 3 ∗
1 2
2 1−𝑦
(3) ,
𝑥
where 𝑦 = 0, 2 , 3 , … … , 𝑥+1 , … …
2.9 If the random variable X has pdf
𝒙−𝟏
𝒇(𝒙) = { 𝟐 , 𝟏 < 𝒙 < 𝟑
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Find a monotone function u(x) such that the random variable 𝒀 = 𝒖(𝑿) has a
uniform (0,1) distribution.
From the probability integral transformation, Theorem 2.1.10, we know that if 𝑢(𝑥) =
𝐹𝑋 (𝑥), then u(x)~uniform (0,1). Therefore, for the given pdf, calculate:
2.11 Let X have the standard normal pdf 𝒇𝑿 (𝒙) = (
𝟏
√𝟐𝝅
𝒙𝟐
) 𝒆− 𝟐 .
(a) Find 𝑬𝑿𝟐 directly, and then by using the pdf of 𝒀 = 𝑿𝟐 from example 2.1.7 and
calculate𝑬𝒀.
(b) Find the pdf of 𝒀 = |𝑿|, and find its mean and variance.
𝑥2
a. Using integration by parts with 𝑢 = 𝑥, 𝑎𝑛𝑑 𝑑𝑣 = 𝑥𝑒 − 2 𝑑𝑥.
∞
𝐸𝑋 2 = ∫ 𝑥 2
−∞
1
√2𝜋
=1
𝑥2
𝑒 − 2 𝑑𝑥 =
1
√2𝜋
𝑥2
∞
𝑥2
−
[−𝑥𝑒 − 2 |∞
−∞ + ∫ 𝑒 2 𝑑𝑥 ] =
−∞
1
√2𝜋
∞
𝑥2
∫ 𝑒 − 2 𝑑𝑥
−∞
2
Using example 2.1.7, let 𝑌 = 𝑋 2 , then
𝑓𝑌 (𝑦) =
∞
𝐸𝑌 = ∫ 𝑦 ∗
0
1
[
1
2√𝑦 √2𝜋
1
𝑦
𝑒 −2 +
1
√2𝜋
1
𝑦
𝑒 −2 ] =
𝑦
√2𝜋𝑦
𝑒 −2
𝑦
√2𝜋𝑦
𝑒 − 2 𝑑𝑦 = 1 (𝐷𝑜 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠)
b.
𝑌 = |𝑋|𝑤ℎ𝑒𝑟𝑒 − ∞ < 𝑥 < ∞. Therefore, 0 < 𝑦 < ∞. Then
𝐹𝑌 (𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(|𝑋| ≤ 𝑦) = 𝑃(−𝑦 ≤ 𝑋 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑦) − 𝑃(𝑋 ≤ −𝑦) =
𝐹𝑋 (𝑦) − 𝐹𝑥 (−𝑦).
𝑑
2
𝑇ℎ𝑢𝑠, 𝑓𝑌 (𝑦) = 𝑑𝑦 𝐹𝑌 (𝑦) = 𝑓𝑋 (𝑦) − 𝑓𝑋 (−𝑦) ∗ (−1) = √𝜋 𝑒
−
𝑦2
2
𝑓𝑜𝑟 𝑦 > 0.
𝑦2
∞
∞
2
2 ∞
2
2
𝑦2
𝐸𝑌 = ∫0 𝑦 ∗ √𝜋 𝑒 − 2 𝑑𝑦 = √𝜋 ∫0 𝑒 −𝑢 𝑑𝑢 = √𝜋 [−𝑒 −𝑢 | ] = √𝜋 , 𝑤ℎ𝑒𝑟𝑒 𝑢 = 2 .
0
𝑦2
𝑦2 ∞
𝑦2
∞
∞
2
2
2
𝜋
𝐸𝑌 2 = ∫0 𝑦 2 ∗ √𝜋 𝑒 − 2 𝑑𝑦 = √𝜋 [−𝑦𝑒 − 2 | + ∫0 𝑒 − 2 𝑑𝑦]= √𝜋 ∗ √ 2 = 1
0
2
𝑉𝑎𝑟(𝑌) = 𝐸𝑌 2 − (𝐸𝑌)2 = 1 − 𝜋
2.13 Consider a sequence of independent coin flips, each of which has probability 𝒑
of being heads. Define a random variable 𝑿 as the length of the run (of either heads
or tails) started by the first trial. (For example, X=3 if either TTTH or HHHT is
observed.) Find the distribution of X, and find EX.
𝑃(𝑋 = 𝑘) = (1 − 𝑝)𝑘 𝑝 + 𝑝𝑘 (1 − 𝑝), 𝑘 = 1,2,3 … Therefore,
∞
𝑘−1
𝑘−1 )
𝐸𝑋 = ∑∞
+ ∑∞
=
𝑘=1 𝑘 ∗ 𝑃(𝑋 = 𝑘) = (1 − 𝑝)𝑝(∑𝑘=1 𝑘(1 − 𝑝)
𝑘=1 𝑘𝑝
(1 − 𝑝)𝑝 [− ∑∞
𝑘=1
𝑑(1−𝑝)𝑘
𝑑𝑝
+ ∑∞
𝑘=1
𝑑𝑝𝑘
1
1
] = (1 − 𝑝)𝑝 (𝑝2 + (1−𝑝)2 ) =
𝑑𝑝
1−2𝑝+2𝑝2
𝑝(1−𝑝)
2.18 Show that if X is continuous random variable, then
𝐦𝐢𝐧 𝑬|𝑿 − 𝒂 |= 𝑬|𝑿 − 𝒎|
𝒂
, where m is the median of X.
∞
𝑎
∞
𝐸|𝑋 − 𝑎| = ∫−∞|𝑥 − 𝑎|𝑓(𝑥)𝑑𝑥 = ∫−∞ −(𝑥 − 𝑎)𝑓(𝑥)𝑑𝑥 + ∫𝑎 (𝑥 − 𝑎)𝑓(𝑥)𝑑𝑥 . Then,
3
𝑎
𝑑
∞
𝐸|𝑋 − 𝑎| = ∫−∞ 𝑓(𝑥)𝑑𝑥 − ∫𝑎 𝑓(𝑥)𝑑𝑥 = 0 , the solution is a=median. This is a
𝑑𝑎
minimum since
𝑑2
𝑑𝑎2
𝐸|𝑋 − 𝑎| = 2𝑓(𝑎) > 0.
2.34 A distribution cannot be uniquely determined by a finite collection of moments,
as this example from Romano and Siegel (1986) shows. Let X have the normal
distribution that is X has pdf
𝒇𝑿 (𝒙) =
𝟏
√𝟐𝝅
𝒙𝟐
𝒆− 𝟐 , −∞ < 𝒙 < ∞
Define a discrete random variable Y by
𝟏
𝟐
𝑷(𝒀 = √𝟑) = 𝑷(𝒀 = −√𝟑) = 𝟔, 𝑷(𝒀 = 𝟎) = 𝟑.
Show that 𝑬𝑿𝒓 = 𝑬𝒀𝒓 for𝒓 = 𝟏, 𝟐, 𝟑, 𝟒, 𝟓.
(Romano and Siegel point out that for any finite n there exists a discrete, and hen
nonnormal, random variable whose first n moments are equal to those of X.
𝑟
𝐸𝑋 =
∞
∫−∞ 𝑥 𝑟
1
∗
𝑟
1
√2𝜋
𝑒
(−
𝑥2
)
2
𝑑𝑥. Thus, 𝐸𝑋 𝑟 = 0, 𝑤ℎ𝑒𝑛 𝑟 = 1,3,5.
𝑟
1
𝐸𝑌 𝑟 = 6 ∗ 32 + 6 ∗ (−1)𝑟 ∗ 32 . Thus, 𝐸𝑋 𝑟 = 0, 𝑤ℎ𝑒𝑛 𝑟 = 1,3,5.
𝑡2
2
𝑑2
𝑑4
𝑀𝑋 (𝑡) = 𝑒 , 𝐸𝑋 2 = 𝑑𝑡 2 𝑀𝑋 (𝑡)| 𝑡=0 = 1, 𝐸𝑋 4 = 𝑑𝑡 4 𝑀𝑋 (𝑡)| 𝑡=0 = 3.
1
2
1
2
4
1
1
4
𝐸𝑌 2 = 6 (√3) + 6 (−√3) = 1, 𝐸𝑌 4 = 6 (√3) + 6 (−√3) = 3.
Thus, 𝐸𝑋 𝑟 = 𝐸𝑌 𝑟 , 𝑓𝑜𝑟 𝑟 = 1,2,3,4,5.
2.38 Let X have the negative binomial distribution with pmf
𝒇𝑿 (𝒙) = (
𝒓+𝒙−𝟏 𝒓
) 𝒑 (𝟏 − 𝒑)𝒙 , 𝒙 = 𝟎, 𝟏, 𝟐, …,
𝒙
, where 0<p<1, and r>0 is an integer.
(a) Calculate the mgf of X.
(b) Define a new random variable by 𝒀 = 𝟐𝒑𝑿. Show that as 𝒑 → 𝟎+ , the mgf of Y
converges to that of a chi squared random variable with 2r degrees of freedom by
1
𝑟
1
showing that lim 𝑀𝑌 (𝑡) = (1−2𝑡) , |𝑡| < 2.
𝑝→0+
4
𝑟+𝑥−1
𝑟+𝑥−1
∞
𝑡𝑥
𝑟
𝑥
𝑟
𝑡 𝑥
a. 𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) = ∑∞
𝑥=0 𝑒 ∗ ( 𝑥 ) 𝑝 (1 − 𝑝) = ∑𝑥=0( 𝑥 ) 𝑝 [(1 − 𝑝)𝑒 ]
𝑟
=
𝑟+𝑥−1 𝑟
𝑡
∑∞
𝑥=0( 𝑥 )𝑝 [1−(1−𝑝)𝑒 ]
[1−(1−𝑝)𝑒 𝑡 ]𝑟
[(1 − 𝑝)𝑒 𝑡 ]𝑥
𝑝𝑟
𝑟+𝑥−1
𝑡 𝑟
𝑡 𝑥
= [1−(1−𝑝)𝑒 𝑡]𝑟 ∑∞
𝑥=0( 𝑥 ) [1 − (1 − 𝑝)𝑒 ] [(1 − 𝑝)𝑒 ]
𝑝𝑟
= [1−(1−𝑝)𝑒 𝑡]𝑟
′
b. 𝑀𝑌 (𝑡) = 𝐸(𝑒 𝑡𝑌 ) = 𝐸(𝑒 2𝑝𝑡𝑋 ) = 𝐸(𝑒 𝑡 𝑋 ), 𝑤ℎ𝑒𝑟𝑒 𝑡 ′ = 2𝑝𝑡.
𝑝𝑟
′
𝐸(𝑒 𝑡 𝑋 ) =
lim
𝑝→0+
𝑝
′ 𝑟
[1−(1−𝑝)𝑒 𝑡 ]
𝑝
1−(1−𝑝)𝑒 2𝑝𝑡
𝑟
= (1−(1−𝑝)𝑒 2𝑝𝑡 ) ,
1
= 1−2𝑡 , by L'Hôpital's rule.
1
𝑟
lim 𝑀𝑌 (𝑡) = (1−2𝑡) .
𝑝→0+
5